I need to measure the reistance of a welded joint.The value of the joint would be in the range of 10 ~ 14 milliohms.My present system has a Constant current source of 500 mAmps at 12 Volts.When I apply the current source to the measuring points the Voltage drop occurs.This millivolt is then fed to an amplifier.Can someone guide me to build the constant current source.Thanks in advance

There are several weys to go about this. Here is a popular basic one. Use a 2.5Ω resistor.

You will need to heat sink the LM317, it will get hot when providing current. Vin isn't critical, though you can't exceed 35V, I'd recomment 12V. The parts are available from Radio Shack, though you might need to put 4 10Ω ½W resistors in parallel.

There are transistor versions, but this one is cheap and easy. Soon as I find the schematic I drew I'll show a transistor schematic.

The easiest way to build a constant current source in that current range is to use an LM317 adjustable positive regulator with a resistor to set the current output.

You connect a resistor from the output terminal to the ADJ terminal, and connect the current source (your +12v) to the input terminal. The constant current is taken from the ADJ terminal.

To get a reasonably accurate current output, you first need to determine the Vref of the regulator. They're nominally 1.25v, but can vary from a low of 1.2v to 1.3v. Vref is read between the output and ADJ terminals when there is a resistor connected between the two and a current path to ground (or supply return) from the ADJ terminal, and current supplied to the input terminal. I suggest using a resistor from 33 to 100 Ohms for this measurement. You should have a heat sink on the LM317.

Once you determine the actual Vref, your output current can be set by:
Iout=Vref/R1
Conversely,
R1=Vref/Iout
So, if Vref=1.25 and desired Iout=0.5A, then R1=1.25/0.5=2.5 Ohms.
Note that nearly all of the circuit current will be flowing through R1, and about 50uA will be supplied via the ADJ terminal; but in this case it's so insignificant that it is negligible.

Note that the power dissipation in the resistor will be 0.625W, or 625mW. For longevity, use a resistor with a much higher power rating. A 2W resistor would have > 3x the rating necessary, and would be a good choice.
[eta]
Yeesh, I'm slow today.

This is because you are using an inefficient setup. If you are driving 0.01-0.015 Ohms with 500mA, you only need (V = I*R) V = 0.5A * 0.015 Ohms = 7.5mV.

So, at 12V, you are dropping nearly all of it across the LM317, let alone what you are giving the LM7812. The power lost is P = V * I , so ~12V * 0.5A, 6W. A good heatsink will rise about 25 degrees celcius from ambient with this sort of load.

You should be using a much lower voltage setup and rethinking the current source approach. I believe using a precision low-input offset opamp will give you your best bet. This, combined with a low-side N-FET or NPN transistor will give you the best performance.

I tried with the one as you described here.But my regulator 7812 and LM 317 get heated up when measuring 10 to 15 milliohms.I have taken ouput from out terminal.Has it to be taken from Adj., terminal?

Click to expand...

The 7812 will also need a heat sink.

You do not source the current from the OUT terminal of the LM317; you source it from the ADJ terminal with the 2.5 Ohm resistor connected from the OUT to the ADJ terminal. If you are attempting to source current directly from the OUT terminal, then your regulators will be supplying the maximum current they can, and will rapidly overheat.

Note that 2.5 Ohms is a nominal value. If your LM317 regulators' Vref is 1.25v, then 2.5 Ohms is the correct resistor to use.
If Vref = 1.2v, then 2.4 Ohms is the correct value to use.
If Vref = 1.3v, then 2.6 Ohms is the correct value to use.

A 7812 regulator has a nominal 2v dropout voltage.
An LM317 regulator used as a current source will have between 2.9v and 3v dropout, largely depending upon Vref.
Therefore, you will have a nominal 5v dropout due to the regulators used.

You might consider converting an ATX-form factor computer power supply to a bench supply. Google "ATX bench supply" for lots of ideas.

That way, you could use just the LM317 as a current regulator from either the +3.3 or +5v supply.

For simplicity, stick with the LM317 configured as a current source. Use an ATX supply with 3.3V to 5V as Wookie suggested. Then, if you do not already have, use a differential amplifier across the measured resistance. Clearly, for this low signal levels, you should be running bipolar supplied low-offset opamps.

Be careful of your 2.5 ohm resistor. Ensure it can handle many times more power than ~1.25*0.5A = 0.625W. This is because the resistor will self-heat. Also, your precision is partially governed by the tolerance on this resistor.

I think a better way would be to pulse a large amount of current to maximize signal to noise ratio without heating anything up. You would need to use a microcontroller to handle the timing and, you would probably have to integrate the current and voltage to find your average resistance. This term would include inductance, so you would have to be careful with that. I guess it really depends on what kind of precision you really need.

I didn't see the 2.5 Ohms inline, which is dropping 1.25V. So, 12V source - 1.25V - very small voltage from load = 10.75V across the LM317. So, this is actually 5.375W across the regulator.

Steve

Click to expand...

Hi Steve,
Well, the regulator tries to keep Vref somewhere between 1.2v and 1.3v, depending upon the individual regulator. TThe "standard" dropout of the regulator is around 1.7v; but that will vary depending upon the amount of current flowing through the regulator. Oddly enough, with currents below 1A, as the temperature of the regulator rises, dropout voltage decreases.

But this still means that the voltage across the regulator and the resistor will be a minimum of around 3v.

This is why I suggested in my most recent prior post to try using the 3.3v or 5v output of an ATX form-factor power supply for the current source. Even at 5v with a load of 0 Ohms, the worst case power dissipation of the LM317 will be 2.5W.

Be careful of your 2.5 ohm resistor. Ensure it can handle many times more power than ~1.25*0.5A = 0.625W. This is because the resistor will self-heat. Also, your precision is partially governed by the tolerance on this resistor.

Click to expand...

The 4 10Ω ½W in parallel I suggested would cover that nicely, for a total of 2W. I believe they come in packs of 4 (5?) from Radio Shack. As for power supplies, 3 D cells in series should work (for a total of 4.5VDC), unless the pipes are dropping more than I think they are.

I don't see any way around it though, there will be some heat generated, probably around the LM317. You can move the heat around with extra resistors, but it just increases complexity. I don't think it is worth the effort overall. With something like D cells the whole package can be built in a small box.